The answer is:
The first triangle is:
[tex]A=59.6\°\\C=71.4\°\\c=17.6units[/tex]
The second triangle is:
[tex]A=120.4\°\\C=10.6\°\\c=3.41units[/tex]
To solve the triangles, we must remember the Law of Sines form.
Law of Sines can be expressed by the following relationship:
[tex]\frac{a}{Sin(A)}=\frac{b}{Sin(B)}=\frac{c}{Sin(C)}[/tex]
Where,
a, b, and c are sides of the triangle
A, B, and C are angles of the triangle.
We are given,
[tex]B=49\°\\a=16\\b=14[/tex]
So, solving the triangles, we have:
Finding A, we have:
[tex]\frac{a}{Sin(A)}=\frac{b}{Sin(B)}\\\\Sin(A)=a*\frac{Sin(B)}{b}=16*\frac{Sin(49\°)}{14}\\\\Sin^{-1}(Sin(A)=Sin^{-1}(16*\frac{Sin(49\°)}{14})\\\\A=59.6\°[/tex]
Finding C, we have:
Now, if the sum of all the interior angles of a triangle is equal to 180°, we have:
[tex]A+B+C=180\°\\\\C=180-A-B\\\\C=180\°-59.6\°-49\°=71.4\°[/tex]
Finding c, we have:
Then, now that we know C, we need to look for "c":
[tex]\frac{14}{Sin(49\°)}=\frac{c}{Sin(71.4\°)}\\\\c=\frac{14}{Sin(49\°)}*Sin(71.4\°)=17.58=17.6units[/tex]
So, the first triangle is:
[tex]A=59.6\°\\C=71.4\°\\c=17.6units[/tex]
Finding A, we have:
[tex]\frac{a}{Sin(A)}=\frac{b}{Sin(B)}\\\\Sin(A)=a*\frac{Sin(B)}{b}=16*\frac{Sin(49\°)}{14}\\\\Sin^{-1}(Sin(A)=Sin^{-1}(16*\frac{Sin(49\°)}{14})\\\\A=59.6\°[/tex]
Now, since that there are two triangles that can be formed, (angle and its suplementary angle) there are two possible values for A, and we have:
[tex]A=180\°-59.6\°=120.4\°[/tex]
Finding C, we have:
Then, if the sum of all the interior angles of a triangle is equal to 180°, we have:
[tex]A+B+C=180\°\\\\C=180\°-A-B\\\\C=180\°-120.4\°-49\°=10.6\°[/tex]
Then, now that we know C, we need to look for "c".
Finding c, we have:
[tex]\frac{14}{Sin(49\°)}=\frac{c}{Sin(10.6)}\\\\c=\frac{14}{Sin(49)\°}*Sin(10.6\°)=3.41units[/tex]
so, The second triangle is:
[tex]A=120.4\°\\C=10.6\°\\c=3.41units[/tex]
Have a nice day!
